Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $t = \dfrac{2(2a - 5)}{-7} \times \dfrac{3a}{6a^2 - 15a} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 2(2a - 5) \times 3a } { -7 \times (6a^2 - 15a) } $ $ t = \dfrac {3a \times 2(2a - 5)} {-7 \times 3a(2a - 5)} $ $ t = \dfrac{6a(2a - 5)}{-21a(2a - 5)} $ We can cancel the $2a - 5$ so long as $2a - 5 \neq 0$ Therefore $a \neq \dfrac{5}{2}$ $t = \dfrac{6a \cancel{(2a - 5})}{-21a \cancel{(2a - 5)}} = -\dfrac{6a}{21a} = -\dfrac{2}{7} $